Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:

import java.util.Scanner;

/*
 * 3
 * 0.30 0.60 0.80
 * */
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        double[] a = new double[n];

        for (int i = 0; i < n; i++) {
            a[i] = sc.nextDouble();
        }
        Double[][] dp = new Double[n][(n + 1) / 2+1];
        System.out.println(coinsMajorityHeadProbablity(a, (n + 1) / 2, dp, 0));
    }

    private static Double coinsMajorityHeadProbablity(double[] a, int x, Double[][] dp, int i) {
        if (x == 0) return 1.0;
        if (i == a.length) return 0.0;

        if (dp[i][x] != null) return dp[i][x];
        double res = a[i] * coinsMajorityHeadProbablity(a, x - 1, dp, i + 1)
                + (1 - a[i]) * coinsMajorityHeadProbablity(a, x, dp, i + 1);
        dp[i][x] = res;
        return res;
    }
}